Which atom bond in atomic interaction combines electrons, filling its valence zone a) Van der Vaals bond; b) a covalent bond; c) ionic bond; d) metallic bond.

True.

Dielectric is a material with a low electrical conductivity (σ << 1); that is, an insulator, which has the property of forming electric dipoles inside it under the action of an electric field. Thus, all dielectric materials are insulators but not all insulating materials are dielectric.

Some examples of this type of materials are glass, ceramics, rubber, mica, wax, paper, dry wood, porcelain, some fats for industrial and electronic use and bakelite. As for the gases, the air, nitrogen and sulfur hexafluoride are used as dielectrics.

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

[tex]P\propto \frac{1}{V}[/tex]     (At constant temperature and number of moles)

[tex]P_1V_1=P_2V_2[/tex]  

where,

[tex]P_1[/tex] = initial pressure of gas  = 150 kPa

[tex]P_2[/tex] = final pressure of gas  = ?

[tex]V_1[/tex] = initial volume of gas   = v L

[tex]V_2[/tex] = final volume of gas  = [tex]\frac{v}{3}L[/tex]

[tex]150\times v=P_2\times \frac{v}{3}[/tex]  

[tex]P_2=450kPa[/tex]

Therefore, the new pressure of the gas will be 450 kPa.

Answer:

(a)[tex]A_1=0.26 m^2[/tex]

(b)Q= -35.69 KW

Explanation:

Given:

[tex]P_1=350 KPa,T_1=420 K,V_1=3 m/s,T_2=300 K,V_2=460 m/s[/tex]

We know that foe air [tex]C_p=1.011\frac{KJ}{kg-k}[/tex]

Mass flow rate for air =2.3 kg/s

(a)

By mass balancing [tex]\dot{m}=\dot{m_1}\dot{m_2}[/tex]

[tex]\dot{m}=\rho AV[/tex]

[tex]\rho_1A_1V_1=\rho_2A_2V_2[/tex]

[tex]\rho_1 =\dfrac {P_1}{RT_1},R=0.287\frac{KJ}{kg-K}[/tex]

[tex]\rho_1 =\dfrac {350}{0.287\times 420}[/tex]

[tex]\rho_1=2.9\frac{kg}{m^3}[/tex]

[tex]\dot{m}=\rho_1 A_1V_1[/tex]

[tex]2.3=2.9\times A_1\times 3[/tex]

[tex]A_1=0.26 m^2[/tex]

(b)

Now from first law for open(nozzle) system

[tex]h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}[/tex]

Δh=[tex]C_p(T_2-T_1)\frac{KJ}{kg}[/tex]

[tex]1.011\times 420+\dfrac{3^2}{2000}+Q=1.011\times 300+\dfrac{460^2}{2000}[/tex]

Q=-15.52 KJ/s

⇒[tex]Q= -15.52\times 2.3[/tex] KW

  Q= -35.69 KW

If heat will loss from the system then we will take negative and if heat will incoming to the system we will take as positive.

Answer:

A) A1 ==0.2829 m^2

B) [tex]\frac{dQ}{dt} = -105.5 kW[/tex]

Explanation:

A) we know from continuity equation

[tex]\frac{dm}{dt} = \frac{A_1 v_1}{V_1}[/tex]

solving for A1

[tex]A_1 = \frac{\frac{dm}{dt} V_1}{v_1}[/tex]

we know V = \frac{RT}{P}  as per ideal gas equation, so we have

[tex]A_1 = = \frac{\frac{dm}{dt} \frac{RT_1}{P_1}}{v_1}[/tex]

       [tex]= \frac{2.3 \frac{0.287 \times 450}{350}}{3}[/tex]

        =0.2829 m^2

b) the energy balanced equation is

[tex]\frac{dQ}{dt} = \frac{dm}{dt} ( Cp(T_2 -T_1) + \frac{V_2^2 - V_1^2}{2})[/tex]

[tex]= 2.3 ( 1.011(300 - 450) + [\frac{460^2+3^2}{2}])[/tex]

[tex]\frac{dQ}{dt} = -105.5 kW[/tex]

Answer:

Cut-off ratio[tex]\dfrac{V_3}{V_2}=6.15[/tex]

Cxpansion ratio[tex]\dfrac{V_4}{V_3}=3.25[/tex]

The exhaust temperature[tex]T_4=1997.5R[/tex]

Explanation:

Compression ratio CR(r)=20

[tex]\dfrac{V_1}{V_2}=20[/tex]

[tex]P_2=P_3=920 psia[/tex]

[tex]T_1=520 R ,T_{max}=T_3,T_3=3200 R[/tex]

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

[tex]\dfrac{T_2}{T_1}=r^{\gamma -1}[/tex]

[tex]\dfrac{T_2}{520}=20^{1.4 -1}[/tex]

[tex]T_2=1723.28R[/tex]

[tex]\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}[/tex]

[tex]\dfrac{V_3}{V_2}=\dfrac{3200}{520}[/tex]

So cut-off ratio[tex]\dfrac{V_3}{V_2}=6.15[/tex]

[tex]\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}[/tex]

Now putting the values in above equation

[tex]\dfrac20=\dfrac{V_4}{V_3}\times 6.15[/tex]

[tex]\dfrac{V_4}{V_3}=3.25[/tex]

So expansion ratio[tex]\dfrac{V_4}{V_3}=3.25[/tex].

[tex]\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}[/tex]

[tex]\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}[/tex]

[tex]T_4=1997.5R[/tex]

So the exhaust temperature[tex]T_4=1997.5R[/tex]

Answer:

Outside temperature =88.03°C

Explanation:

Conductivity of air-soil from standard table

   K=0.60 W/m-k

To find temperature we need to balance energy

Heat generation=Heat dissipation

Now find the value

We know that for sphere

[tex]q=\dfrac{2\pi DK}{1-\dfrac{D}{4H}}(T_1-T_2)[/tex]

Given that q=500 W

so

[tex]500=\dfrac{2\pi 2\times .6}{1-\dfrac{2}{4\times 10}}(T_1-25)[/tex]

By solving that equation we get

[tex]T_2[/tex]=88.03°C

So outside temperature =88.03°C

Answer:

b) False

Explanation:

In close system only energy transfer take place and mass transfer is zero.But on the other hand in open system energy as well mass transfer take place.

Energy transfer means work as well as heat transfer.But we know that in close system there is no any flow of mass so there will not be any flow work,only boundary work will associated with close system.But in open system flow work take place.

Solution:

Given:

mass of air, m = 0.50 Kg

[tex]T_{1}[/tex] = 25°C = 273+25 = 298 K

[tex]T_{2}[/tex] = 100°C = 273+100 = 373 K

[tex]T_{o}[/tex] = 200°C = 273+100 = 473 K

Solution:

Formulae used:

ΔQ = mCΔT                                          (1)

ΔS = [tex]\frac{\Delta Q}{T_{o}}[/tex]    (2)

where,

ΔQ = change in heat transfer

ΔS = chane in entropy

C = specific heat

ΔT = change in system temperature

Using eqn (1)

ΔQ = [tex]0.50\times 1.005\times (373-298)[/tex] = 36.687 kJ

Now, for entropy generation, using eqn (2)

ΔS = [tex]\frac{37.687}{473}[/tex] = 0.0796 kJ

Answer:

cold working

Explanation:

Given data in question

melting point tungsten (W) = 3380°C = 3653 K

processing temperature = 1100°C = 1373 K

To find out

process is hot or cold working

solution

we know hot working and cold working process is depend upon the Recrystallization and Recrystallization is a process in which deformed grains of crystal structures are replace with stress-free grains that nucleate and grow till actual grains have been consumed fully

and we know that ratio of processing temperature and melting point tungsten is greater than 60% than the process is start of Recrystallization so we check ratio

ratio =  processing temperature / melting point tungsten

ratio =  1373 / 3653

ratio = 0.3758 = 37.58 %

we can see this is less than 60 % so our process is cold working

Answer:

Enthalpy at outlet=284.44 KJ

Explanation:

[tex]m_1=1 Kg/s,m_2=1.5 Kg/s,m_3=22 Kg/s[/tex]

[tex]h_1=100 KJ/Kg,h_2=120 KJ/Kg,h_3=500 KJ/Kg[/tex]

We need to Find enthalpy of outlet.

Lets take the outlet mass m and outlet enthalpy h.

So from mass conservation

[tex]m_1+m_2+m_3=m[/tex]

   m=1+1.5+2 Kg/s

  m=4.5 Kg/s

Now from energy conservation

[tex]m_1h_1+m_2h_2+m_3h_3=mh[/tex]

By putting the values

[tex]1\times 100+1.5\times 120+2\times 500=4.5\times h[/tex]

So h=284.44 KJ

Answer:-0.4199 J/k

Explanation:

Given data

mass of nitrogen(m)=1.329 Kg

Initial pressure[tex]\left ( P_1\right )[/tex]=120KPa

Initial temperature[tex]\left ( T_1\right )=27\degree \approx[/tex] 300k

Final volume is half of initial

R=particular gas constant

Therefore initial volume of gas is given by

PV=mRT

V=0.986\times 10^{-3}

Using [tex]PV^{1.49}[/tex]=constant

[tex]P_{1}V^{1.49}[/tex]=[tex]P_2\left (\frac{V}{2}\right )[/tex]

[tex]P_2[/tex]=337.066KPa

[tex]V_2[/tex]=[tex]0.493\times 10^{-3} m^{3}[/tex]

and entropy is given by

[tex]\Delta s[/tex]=[tex]C_v \ln \left (\frac{P_2}{P_1}\right )[/tex]+[tex]C_p \ln \left (\frac{V_2}{V_1}\right )[/tex]

Where, [tex]C_v[/tex]=[tex]\frac{R}{\gamma-1}[/tex]=0.6059

[tex]C_p[/tex]=[tex]\frac{\gamma R}{\gamma -1}[/tex]=0.9027

Substituting values we get

[tex]\Delta s[/tex]=[tex]0.6059\times\ln \left (\frac{337.066}{120}\right )[/tex]+[tex]0.9027 \ln \left (\frac{1}{2}\right )[/tex]

[tex]\Delta s[/tex]=-0.4199 J/k

Answer:

true area of contact is 1.7 * [tex]10^{-4}[/tex] m²

Explanation:

Given data

mass (m) = 4000 tonnes

yield strength = 240 MPa i.e. = 240 * [tex]10^{6}[/tex]

base area = 0.8 m²

To find out

the true area of contact

Solution

we have given yield strength and weight

so with we can find contact area directly we know that

area is equal to weight / yield strength

so we will put weight and yield strength value in this formula

and weight = mass * 9.81 = 4 * 9.81 = 39.24 tonnes = 39240 N

area = weight /  yield strength

area = 39240 / 240 * [tex]10^{6}[/tex]

true area of contact = 1.7 * [tex]10^{-4}[/tex] m²

Answer:

(a) [tex]m_{R-134a}=0.0338kg/s[/tex]

(b) [tex]Q_H=7.03kW[/tex]

(c) [tex]COP=4.39[/tex]

Explanation:

Hello,

(a) In this part, we must know that the energy provided by the water equals the energy gained by the refrigerant-134a, thus:

[tex]m_{R-134a}(h_2-h1)=m_{H_2O}Cp_{H_2O}*\Delta T_{H_2O}[/tex]

Now, water's heat capacity is about 4.18kJ/kg°C and the enthalpies at both the first and second state for the refrigerant-134a are computed as shown below, considering the first state as a vapor-liquid mixture (VLM) at 12°C and the second state as a saturated vapor (ST) at the same conditions:

[tex]h_{1/12^0C/VLM}=68.18kJ/kg+0.15*189.09kJ/kg=96.54kJ/kg\\h_{2,SV}=257.27kJ/kg[/tex]

Next, solving the mass of water one obtains:

[tex]m_{R-134a}=\frac{m_{H_2O}Cp_{H_2O}*\Delta T_{H_2O}}{(h_2-h1)}=\frac{0.065kg/s*4.18kJ/kg^0C*(60^0C-40^0C)}{257.27kJ/kg-96.54kJ/kg} \\m_{R-134a}=0.0338kg/s[/tex]

(b) Now, the energy balance allows us to compute the heat supply:

[tex]Q_L+W_{in}=Q_H\\Q_H=0.0338kg/s*(257.27kJ/kg-96.54kJ/kg)+1.6kW\\Q_H=7.03kW[/tex]

(c) Finally, the COP (coefficient of performance) is computed via:

[tex]COP=\frac{Q_H}{W_{in}}=\frac{7.03kW}{1.6kW}\\COP=4.39[/tex]

Best regards.

Answer:

b) Endothermic Chemical Reactions in a solid

Explanation:

Endothermic reactions consume energy, which will result in a cooler solid when the reaction finishes.

Answer:

C.The chaotic arrangement of atoms or high hardness

Explanation:

We know that atomic arrangement in Solids are of two types

 1)Crystalline

 2)Amorphous

Crystalline arrangement have periodic arrangement where as Amorphous arrangement have random arrangement.

Generally all metal have Crystalline arrangement and material like wood ,glass have  random arrangement ,that is why wood and glass is called Amorphous.We know that wood act as a insulator for conductivity and glass is a brittle and hard material.

So from above we can say that Amorphous material have chaotic arrangement of atoms or have high harness,so our option c is right.

Elastic deformation is a temporary, reversible change in a material's crystal lattice under stress, following Hooke's law, and is depicted as a linear response on a stress-strain graph. Plastic deformation results in permanent, irreversible changes in the crystal structure, typically involving dislocations and is characterized by the yield point and yield stress. Factors such as temperature and rate of stress application influence a rock's response to stress.

Deformation in materials can be categorized into two types: elastic deformation and plastic deformation. Elastic deformation refers to temporary changes in the crystal lattice that are reversible when the applied stress is removed. It follows Hooke's law, where the force is proportional to the displacement, and is shown as a linear region on a stress-strain graph. On the other hand, plastic deformation results in permanent changes to the lattice structure. Dislocations play a significant role in this process, where planes of atoms slip past one another. The moment when deformation transitions from elastic to plastic is known as the yield point, with associated yield stress. Beyond this point, deformation is irrevocable, and with continued stress, the material will eventually fracture.

At a microscopic level, plastic deformation involves the movement of dislocations and the introduction of an extra plane of atoms in the crystal structure, which allows atoms to move more easily under stress. This results in a permanently altered lattice configuration. Elastic deformation, by contrast, can be envisioned as if atoms were connected by springs that return to their original positions after the removal of stress.

The ability of rocks to deform elastically or plastically before breaking depends on several factors including temperature, water content in clay-bearing rocks, the rate at which stress is applied, and the inherent strength of the rock. A fundamental understanding of these principles is essential in geology and materials science.

Elastic deformation is reversible, maintaining lattice structure. Plastic deformation is irreversible, causing permanent lattice rearrangement.

Elastic and plastic deformation are two different responses of materials to applied stress, and they affect the crystal lattice structure differently:

1. Elastic Deformation :

  - Elastic deformation occurs when a material is subjected to stress, but it returns to its original shape and size once the stress is removed.

  - In elastic deformation, the atomic or molecular bonds within the crystal lattice are stretched or compressed, causing the material to temporarily change shape.

  - Within the elastic limit, the crystal lattice structure remains intact, and the atoms or molecules maintain their relative positions.

  - The deformation is reversible, meaning the material returns to its original state when the applied stress is released.

2. Plastic Deformation :

  - Plastic deformation occurs when a material is subjected to stress beyond its elastic limit, causing permanent changes in shape or size even after the stress is removed.

  - In plastic deformation, the atomic or molecular bonds within the crystal lattice undergo significant rearrangement or sliding.

  - Plastic deformation leads to the permanent displacement of atoms or molecules within the lattice structure, resulting in the material maintaining a new shape or size.

  - The material undergoes irreversible changes in its crystal lattice structure due to dislocation movement, grain boundary sliding, or other mechanisms.

  - Plastic deformation is characteristic of materials undergoing permanent deformation, such as metals being shaped or formed through processes like forging, rolling, or extrusion.

In summary, the difference between elastic and plastic deformation lies in the extent of the changes to the crystal lattice structure and whether the deformation is reversible or permanent. Elastic deformation involves temporary changes within the elastic limit, whereas plastic deformation involves permanent changes beyond the elastic limit.

Answer:

n=2.32

w= -213.9 KW

Explanation:

[tex]V_1=0.3m^3,T_1=298 K[/tex]

[tex]V_2=0.1m^3,T_1=1273 K[/tex]

Mass of air=1 kg

For polytropic process  [tex]pv^n=C[/tex] ,n is the polytropic constant.

  [tex]Tv^{n-1}=C[/tex]

  [tex]T_1v^{n-1}_1=T_2v^{n-1}_2[/tex]

[tex]298\times .3^{n-1}_1=1273\times .1^{n-1}_2[/tex]

n=2.32

Work in polytropic process given as

       w=[tex]\dfrac{P_1V_1-P_2V_2}{n-1}[/tex]

      w=[tex]mR\dfrac{T_1-T_2}{n-1}[/tex]

Now by putting the values

w=[tex]1\times 0.287\dfrac{289-1273}{2.32-1}[/tex]

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

  We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

Answer: B- Nitrogen Tetroxide

Explanation: Except for the nitrogen tetroxide , other given all options are fuel .Nitrogen Tetroxide is a chemical compound having brownish-red color which is in liquid form having a unpleasant smell, therefore it does not belong to the category of fuel because it cannot be used as a substance for production of heat or power .

Answer:

55.655 lb/ft³

Explanation:

Given data in question

oil weight i.e. w  = 350 lb    

oil volume i.e. v = 65 gallons = 6.68403 ft³

To find out

the specific weight of the oil

Solution

We know the specific weight formula is weight / volume    

we have given both value so we will put weight and volume value in

specific weight formula i.e.  

specific weight  =  weight / volume    

specific weight  =  372 / 6.68403 = 55.6550    

specific weight  =  55.655 lb/ft³

By increasing magnification you decrease the field of view.

The answer is A.

Hope this helps.

r3t40

Answer:

w =  -28.8 kJ/kg

q = 723.13 kJ/kg

Explanation:

Given :

Initial properties of piston  cylinder assemblies

Temperature, [tex]T_{1}[/tex] = -20°C

Quality, x = 70%

           = 0.7

Final properties of piston  cylinder assemblies

Temperature, [tex]T_{2}[/tex] = 180°C

Pressure, [tex]P_{2}[/tex] = 6 bar

From saturated ammonia tables at [tex]T_{1}[/tex] = -20°C  we get

[tex]P_{1}[/tex] = [tex]P_{sat}[/tex] = 1.9019 bar

[tex]v_{f}[/tex] = 0.001504 [tex]m^{3}[/tex] / kg

[tex]v_{g}[/tex] = 0.62334 [tex]m^{3}[/tex] / kg

[tex]u_{f}[/tex] = 88.76 kJ/kg

[tex]u_{g}[/tex] = 1299.5 kJ/kg

Therefore, for initial state 1 we can find

[tex]v_{1}[/tex] = [tex]v_{f}[/tex]+x ([tex]v_{g}[/tex]-[tex]v_{f}[/tex]

                       = 0.001504+0.7(0.62334-0.001504)

                       = 0.43678 [tex]m^{3}[/tex] / kg

[tex]u_{1}[/tex] = [tex]u_{f}[/tex]+x ([tex]u_{g}[/tex]-[tex]u_{f}[/tex]

                       = 88.76+0.7(1299.5-88.76)

                       =936.27 kJ/kg

Now, from super heated ammonia at 180°C, we get,

[tex]v_{2}[/tex] = 0.3639 [tex]m^{3}[/tex] / kg

[tex]u_{2}[/tex] = 1688.22 kJ/kg

Therefore, work done, W = area under the curve

           [tex]w = \left (\frac{P_{1}+P_{2}}{2}  \right )\left ( v_{2}-v_{1} \right )[/tex]

           [tex]w = \left (\frac{1.9019+6\times 10^{5}}{2} \right )\left ( 0.3639-0.43678\right )[/tex]

           [tex]w = -28794.52[/tex] J/kg

                       = -28.8 kJ/kg

Now for heat transfer

[tex]q = (u_{2}-u_{1})+w[/tex]

[tex]q = (1688.2-936.27)-28.8[/tex]

          = 723.13 kJ/kg

Answer:

15.8529 kW

Explanation:

Rate of heat loss = 60000 Btu/h

Internal heat gain = 6000 Btu/h

Rate of heat required to be supplied

[tex]P_{Sup}=\text{Rate of heat loss}-\text{Internal heat gain}\\\Rightarrow P_{Sup}=60000-6000\\\Rightarrow P_{Sup}=54000\ Btu/h[/tex]

Converting 54000 Btu/h to kW (kJ/s)

1 Btu = 1.05506 kJ

1 h = 3600 s

[tex]P_{Sup}=54000\times \frac{1.05506}{3600}\\\Rightarrow P_{Sup}=15.8529\ kW[/tex]

∴ Required rated power of these heaters is 15.8529 kW

Answer:

Q = 15.8 kW

Explanation:

Given data:

Heat loss rate is 60,000 Btu/h

Heat gain is 6000 Btu/h

Rate of heat required is computed as

Q = (60000 - 6000) Btu/h

Q = 54000 Btu/h

change Btu/h to Kilo Watts

[tex]Q = 54000 Btu/h (\frac{1W}{3.412142\ Btu/h})[/tex]

[tex]Q = 15825.8 W(\frac{1 kW}{1000 W})[/tex]

Q = 15.8 kW

Answer:

Cooling Rate=340 W

Coefficient of Performance β=3.4

Explanation:

[tex]Desired\ effect= Cooling\ Rate=Q_L= 440-100=340\ W\\ W_{net,in}=Work\ in=100\ W[/tex]

[tex]Coefficient\ of\ performance (\beta) =\frac {Desired\ Out}{Required\ In}=\frac {Cooling\ {Effect}}{Work\ In}=\frac {Q_L}{W_{net,in}}[/tex]

[tex]Coefficient\ of\ performance (\beta) =\frac {440-100}{100}=3.4[/tex]

Answer:[tex]\dot{m}=3.46lbm/min[/tex]

Explanation:

Initial conditions

[tex]P_1=15 psia[/tex]

[tex]T_1=60 F^{\circ}[/tex]

Final conditions

[tex]P_2=75 psia[/tex]

[tex]T_2=400F^{\circ}[/tex]

Steady flow energy equation

[tex]\dot{m}\left [ h_1+\frac{v_1^2}{2}+gz_1\right ]+\dot{Q}=\dot{m}\left [ h_2+[tex]\frac{v_2^2}{2}+gz_2\right ]+\dot{W}[/tex]

[tex]\dot{m}\left [ c_pT_1+\frac{0^2}{2}+g0\right ]+\dot{Q}=\dot{m}\left [ c_pT_2+\frac{0^2}{2}+g0\right ]+\dot{W}[/tex]

[tex]\dot{m}c_p\left [ T_1-T_2\right ]+\left [ -5hp\right ]=\dot{W} -5\times 746\times 3.4121[/tex]

[tex]-4\dot{m}-\dot{m}\times 0.24\times \left [ 400-60\right ][/tex]

[tex]-81.6\dot{m}-4\dot{m}=-4.949 BTU/sec[/tex]

[tex]\dot{m}=0.057821lbm/sec[/tex]

[tex]\dot{m}=3.46lbm/min[/tex]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 [tex]h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}[/tex]

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

[tex]Total\ heat\ =C_p\Delta T+\Delta h[/tex]

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

  [tex]h_1=\dfrac{\Delta h}{\Delta T}[/tex]

  [tex]h_1=\dfrac{2406}{20}[/tex]

[tex]h_1=120.3\frac{KJ}{kg-m^2K}[/tex]

Now film coefficient after inclusion of sensible heat

 [tex]h_2=\dfrac{total\ heat}{\Delta T}[/tex]

 [tex]h_2=\dfrac{2,544}{20}[/tex]

[tex]h_2=127.2\frac{KJ}{kg-m^2K}[/tex]

[tex]So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100[/tex]

             [tex]=\dfrac{127.2-120.3}{120.3}\times 100[/tex]

                   =5.75 %

So Percentage change 5.75 %.

Answer:

Explanation:

Thermodynamics properties are the properties which defined the state of any system.

some of the thermodynamics properties are pressure, temperature etc

thermodynamics are broadly divided into two type

1)intensive and

2)extensive properties

Dependent properties are the properties that are dependent on other properties. Extensive property are those which are dependent on the extent of system. Example volume. if size of the system increase or decrease then volume also have same effect according to the changes

Answer:

Explanation:

Low pressure die casting -

Also called the cold chamber die casting .

Example is -

A380 - having the composition , Al ( > 80% ) , Cu( 3 - 4% ) , Si ( 7.5 - 9.5% )

High Pressure die casting -

Also called hot chamber die casting .

Example is -  

ZAMAK 2 - having composition , Al ( 3.5 - 4.3% ) , Cu ( 2.5 - 3.5% ) , Zn( > 90% )

Low pressure die casting -

This type of die casting is perfect for the metals with  high melting point , for example aluminium . during this process , the metal is liquefied by very high temperature in the furnace  and then loaded in to the cold chamber to be injected to the die.

High Pressure die casting -

The metal is melted in a container and then a piston injects the liquid metal under high pressure into the die . low melting point metals that don not chemically attack are ideal for this die casting , example Zinc.

Answer:

 Redundant work refers to the work done during the process of deformation due to friction. It happens during the wire drawing. Redundant work per unit volume increases when the radial position becomes higher. The redundant work factor is defined as increased strain of the deformation to the stress. It is basically related to the deformation area geometry.

Answer: C) actuator

Explanation:

Actuator is the device that used to provides the power and manipulate the motion of the moving parts of the valve and damper is used to control the flow of the fluid. Actuator is the device or the mechanism which are used to control valve automatically and valve is a device which is used to control and regulate the fluid by rotating the flow.

Answer:The most advantage of fuel cells is that can produce electrical energy directly from chemical energy of hydrogen or other fuel.

Explanation: Fuel cell utilizes the chemical energy from the hydrogen or any other fuel and then converts it to the electrical energy. A fuel like hydrogen is supplied to the anode part and air is supplied to the cathode part . For hydrogen fuel cell there is a catalyst at anode side which divides hydrogen molecules in protons and electrons, which split and take go in different direction to cathode side. Thus the fuel cell works and generate the electrical energy